Question

Find the coordinates of the point
divided which trisect the points
(1,-2) and (-3, 4)
3
Show Your Work​

Answer 1

Answer

The coordinates of the points trisects the line AB are either (-1/3 , 0) or ( -5/3 , 0)

Given

The coordinates of the points are (1 , -2) and (-3 , 4)

A point trisects these given points i.e. divides into 3 equal parts

To Find

The coordinates of the points dividing the given lines

Solution

Given , the coordinates of the points A(1 , -2) and B(-3 , 4)

Let us consider a point C(x , y) divides the line AB into three equal parts ot trisects .

Therefore , the probable ratios will be either 1:2 or 2:1

Taking ratio as 1:2

$\sf x = \dfrac{1 \times ( - 3) + 2 \times 1}{1 + 2} \\ \\ \sf \implies x = \dfrac{ - 3+ 2}{3} \\ \\ \sf \implies x = \dfrac{ - 1}{3}$

And

$\sf y = \dfrac{1 \times (4) + 2 \times ( -2)}{1 + 2} \\ \\ \sf \implies y = \dfrac{4 - 4}{3} \\ \\ \sf \implies y = 0$

Therefore , the point (-1/3 , 0) trisects the line AB

Now taking ratio as 2:1

$\sf x = \dfrac{2 \times ( - 3) + 1 \times1}{1 +2} \\ \\ \sf \implies x = \dfrac{ - 6 + 1}{3} \\ \\ \sf \implies x = \dfrac{ - 5}{3}$

And

$\sf y = \dfrac{2\times (4) + 1\times (-2)}{1+2} \\\\ \sf\implies y = \dfrac{8 - 2}{3} \\\\ \sf\implies y = \dfrac{6}{3} \\\\ \sf\implies y = 2$

Therefore , the other point trisects the line AB is (-5/3 , 2)

Answer 2

Let , P and Q are the points which trisect the point A(1,-2) and (-3,4)

We know that , the section formula is given by

$\boxed{ \tt{x = \frac{m x_{2} + n x_{1} }{m + n} \: \: , \: \: y = \frac{m y_{2} + n y_{1} }{m + n} }}$

So , P divides the line segment in ratio 1 : 2

Thus ,

$\tt \implies x = \frac{ - 3(1) + 1(2)}{1 + 2} \: \: , \: \: y = \frac{4(1) + ( - 2)(2)}{1 + 2}$

$\tt \implies x = \frac{ - 3 + 2}{3} \: \: , \: \: y = \frac{4 - 4}{3}$

$\tt \implies x = \frac{ - 1}{3} \: \:, \: \: y = 0$

Now , Q divides the line segment in ratio 2 : 1

Thus ,

$\tt \implies x = \frac{ - 3(2) + 1(1)}{2 + 1} \: \: , \: \: y = \frac{4(2) + ( - 2)(1)}{2 + 1}$

$\tt \implies x = \frac{ - 6 + 1}{ 3} \: \:, \: \: y = \frac{ - 8 + 2}{3}$

$\tt \implies x = \frac{ - 5}{3} \: \:, \: \: y = \frac{ -2}{1}$

Hence , P(-1/3 , 0) and Q(-5/3 , -2) are the points which trisect the line segment AB